我必须使用 IN 运算符从数据库中选择一些行。我想用准备好的语句来做。这是我的代码:
<?php
$lastnames = array('braun', 'piorkowski', 'mason', 'nash');
$in_statement = '"' . implode('", "', $lastnames) . '"'; //"braun", "piorkowski", "mason", "nash"
$data_res = $_DB->prepare('SELECT `id`, `name`, `age` FROM `users` WHERE `lastname` IN (?)');
$data_res->bind_param('s', $in_statement);
$data_res->execute();
$result = $data_res->get_result();
while ($data = $result->fetch_array(MYSQLI_ASSOC)) {
...
}
?>
但是尽管所有数据都存在于数据库中，但什么也没有返回。
还有一个:如果我直接通过$in_statement来查询和执行它，将返回数据.所以问题出现在准备上。
我在谷歌上寻找这个问题，但没有成功。我的代码有什么问题？
感谢您的帮助!
我最近找到了我的问题的解决方案。也许这不是最好的方法，但效果很好!证明我错了:)
<?php
$lastnames = array('braun', 'piorkowski', 'mason', 'nash');
$arParams = array();
foreach($lastnames as $key => $value) //recreate an array with parameters explicitly passing every parameter by reference
$arParams[] = &$lastnames[$key];
$count_params = count($arParams);
$int = str_repeat('i',$count_params); //add type for each variable (i,d,s,b); you can also determine type of the variable automatically (is_int, is_float, is_string) in loop, but i don't need it
array_unshift($arParams,$int);
$q = array_fill(0,$count_params,'?'); //form string of question marks for statement
$params = implode(',',$q);
$data_res = $_DB->prepare('SELECT `id`, `name`, `age` FROM `users` WHERE `lastname` IN ('.$params.')');
call_user_func_array(array($data_res, 'bind_param'), $arParams);
$data_res->execute();
$result = $data_res->get_result();
while ($data = $result->fetch_array(MYSQLI_ASSOC)) {
...
}
$result->free();
$data_res->close();
?>